c++ - 类不一致

Question

我制作了一个似乎有效的 nibble 类，但是当我多次使用它时，结果就不同了。下面的代码应该可以说明问题。

#include <cstdlib>
#include <iostream>
using namespace std;
class nibble{
public:
    nibble(){}
    nibble(int n){
        for (int x=0;x<4;x++){
            b[x]=bool(n%2);
            //cout<< x<<"b[x]="<<b[x]<<endl;
            n/=2;
        }
    }
    nibble(char h){
        char* end;
        int n = strtol(&h,&end,16);
        for (int x=0;x<4;x++){
            b[x]=n%2;
            n/=2;
        }
    }
    bool bit(int n){
        n-=2;// this should only have to be n--
        return b[n];
    }
    void set(int n,bool bl ){
        n-=2;//this should only have to be n-- but n-- doesn't work.
        b[(n)]=bl;
    }
    string bin(){
        string out;
        if (b[0]){out+='1';}
        else{out+='0';}
        if (b[1]){out+='1';}
        else{out+='0';}
        if (b[2]){out+='1';}
        else{out+='0';}
        if (b[3]){out+='1';}
        else{out+='0';}

        out+='b';//cout<<'b'<<endl;;
        return out;
    }
    char hex(){
        int out=0;
        for (int x=3; x>-1; x--){
            out *=2;
            out+=b[x];
        }
        //cout<< "hex() out="<<out<<endl;
        switch (out){
            case 0:
            return '0';
            break;  
            case 1:
            return '1';
            break;  
            case 2:
            return '2';
            break;  
            case 3:
            return '3';
            break;  
            case 4:
            return '4';
            break;  
            case 5:
            return '5';
            break;  
            case 6:
            return '6';
            break;  
            case 7:
            return '7';
            break;  
            case 8:
            return '8';
            break;  
            case 9:
            return '9';
            break;  
            case 10:
            return 'A';
            break;  
            case 11:
            return 'B';
            break;  
            case 12:
            return 'C';
            break;  
            case 13:
            return 'D';
            break;  
            case 14:
            return 'E';
            break;  
            case 15:
            return 'F';
            break;  
        }
}
    int val(){
        int out=0;
        for (int x=3; x>-1; x--){
            out *=2;
            out+=b[x];
        }
        return out;
    }

private:
    bool b[3];
};
/**
nibble operator~(nibble in);
nibble operator|(nibble in1, nibble in2);
nibble operator&(nibble in1, nibble in2);
*/
int main(){
char c;
nibble a(10);
cout<<a.bin()<<" "<<a.hex()<<" "<<a.val()<<endl;
cout<<a.bit(4)<<' '<<a.bit(3)<<' '<<a.bit(2)<<' '<<a.bit(1)<<'b'<<endl;

/*** when this code is inserted, the above code gives a different result. I have no clue why this happens

cin>>c;
nibble b('A');
cout<<b.bin()<<" "<<b.hex()<<" "<<b.val()<<endl;
cout<<b.bit(4)<<' '<<b.bit(3)<<' '<<b.bit(2)<<' '<<b.bit(1)<<'b'<<endl;/*
***/
}

我觉得这一定很愚蠢，但我已经研究了一段时间，但找不到问题所在。提前致谢。

Answer 1

进行以下更改:

    bool bit(int n){
        return b[n-1];
    }

    void set(int n,bool bl ){
        b[n-1]=bl;
    }

最后(你怎么错过了这个？)

private:
    bool b[4]; //you had b[3]